What is the vertices f(x)=x^2+10x+16 ?

The vertices f(x)=x^2+10x+16 is Minimum (-5,-9)

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vertices f(x)=x^2+10x+16

vertices

f (x) = x^{2} + 10 x + 16

Minimum (- 5, - 9)

Solution steps

y = x^{2} + 10 x + 16

The parabola parameters are:

a = 1, b = 10, c = 16

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{10}{2 \cdot 1}

Simplify

x_{v} = - 5

Plug in

x_{v} = - 5

to find the

y_{v}

value

y_{v} = - 9

Therefore the parabola vertex is

(- 5, - 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 5, - 9)

f (x) = x^{2} - 10 x + 25

f (x) = (x + 2)^{2} - 1

- x^{2} + 10 x

x^{2} + 4 x - 8 y + 12 = 0

y = - \frac{1}{4} x^{2} + 8 x - 4

What is the vertices f(x)=x^2+10x+16 ?
The vertices f(x)=x^2+10x+16 is Minimum (-5,-9)